package acm.第二轮;

import java.util.Scanner;

/**
 * 思路：
 * 1.遍历所有偶数，设当前偶数为i
 * 2.依次判断 2-i/2之内的的质数是否符合要求
 */
public class A {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int i = 4;
        while (i <= n) {
            int[] fun = fun(i);
            if (fun != null) {
                System.out.println(i + "=" + fun[0] + "+" + fun[1]);
            }
            i += 2;
        }
    }

    public static int[] fun(int n) {
        for (int i = 2; i <= n / 2; i++) { //找出2到n/2之内的质数
            if (isPrime(i)&&isPrime(n-i)) {
                return new int[]{i,n-i};
            }
        }
        return null;
    }

    public static boolean isPrime(int n) {
        if (n == 2) {
            return true;
        }
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }
}
